If the points having the position vectors $3 \hat{i}-2 \hat{j}-\hat{k}, 2 \hat{i}+3 \hat{j}-4 \hat{k}, -\hat{i}+\hat{j}+2 \hat{k}$ and $4 \hat{i}+5 \hat{j}+\lambda \hat{k}$ are coplanar,then $\lambda=$

The value of $(\vec{a} + 2\vec{b} - \vec{c}) \cdot \{(\vec{a} - \vec{b}) \times (\vec{a} - \vec{b} - \vec{c})\}$ is equal to

Consider the vectors $\vec{a}=2 \hat{i}+3 \hat{j}-6 \hat{k}$, $\vec{b}=6 \hat{i}-2 \hat{j}+3 \hat{k}$ and $\vec{c}=3 \hat{i}-6 \hat{j}-2 \hat{k}$.
Assertion $(A):$ The three vectors do not form a triangle.
Reason $(R):$ The three vectors are non-coplanar.
The correct option among the following is:

If $x, y$ and $z$ are non-zero real numbers and $\vec{a}=x \hat{i}+2 \hat{j}, \vec{b}=y \hat{j}+3 \hat{k}$ and $\vec{c}=x \hat{i}+y \hat{j}+z \hat{k}$ are such that $\vec{a} \times \vec{b}=z \hat{i}-3 \hat{j}+xy \hat{k}$ is not given,but $\vec{a} \times \vec{b}=6 \hat{i}-3 \hat{j}+\hat{k}$ is given as $z \hat{i}-3 \hat{j}+\hat{k}$,then the scalar triple product $[\vec{a} \vec{b} \vec{c}]$ is equal to:

The four points whose position vectors are given by $2\bar{a}+3\bar{b}-\bar{c}$, $\bar{a}-2\bar{b}+3\bar{c}$, $3\bar{a}+4\bar{b}-2\bar{c}$ and $\bar{a}-6\bar{b}+6\bar{c}$ are

For what value of $a$ is the volume of the parallelepiped formed by the vectors $\hat{i} + a\hat{j} + \hat{k}$,$\hat{j} + a\hat{k}$,and $a\hat{i} + \hat{k}$ minimum?